This is another combinatorics problem. Each seat in the class will be occupied either by a girl or a boy. The number of distinct girl/boy arrangements for a class with 20 seats is 220, or 1,048,576. Among these arrangements, the number with 10 girls and 10 boys is 20! / (10! * 10!), or 184,756. This is “20 choose 10”. The chances are thus about 18%.

The number of arrangements with k girls and 20-k boys follows the binomial distribution.

Beyond the trivia: The above analysis assumes that every seat has an equal probability of being occupied by a girl or a boy, independent of the other seats. As a result, each of the 1,048,576 arrangements is equally likely. The “equally represented” language is a good justification for this assumption. However, if the school population is small, then the assumption will be inaccurate. For instance, suppose there are actually just 20 students in the school, including 10 girls and 10 boys. Then the class will always have 10 of each; other arrangements are impossible. A similar caveat applied when looking at the probability that two friends were seated in the same class.